Indeterminate Equation #1

$x(3y+2)+y=171$ . Multiply $3$ then add $2$ to each side.

$3x(3y+2)+3y+2=515$ . Factorize this.

$(3x+1)(3y+2)=515$ .

$515=5\times103$ . Note that $5\div3$ has a remainder of $2$ , and $103\div3$ has a remainder of $1$ .

Since both $3x+1$ and $3y+2$ are natural numbers, $3x+1=103$ , $3y+2=5$ .

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$\therefore \quad x=34,\quad y=1$

$\therefore \quad x-y=\boxed{33}$ .